If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and ${{37} \over 4}$

respectively, then (a $-$ b)² is equal to :

Mean = ${{6 + 10 + 7 + 13 + a + 12 + b + 12} \over 8} = 9$<br><br>60 + a + b = 72<br><br>a + b = 12 .....(1)<br><br>variance $$ = {{\sum {x_i^2} } \over n} - {\left( {{{\sum {x_i^{}} } \over n}} \right)^2} = {{37} \over 4}$$<br><br>$$\sum {x_i^2} = {6^2} + {10^2} + {7^2} + {13^2} + {a^2} + {b^2} + {12^2} + {12^2} = {a^2} + {b^2} + 642$$<br><br>${{{a^2} + {b^2} + 642} \over 8} - {(9)^2} = {{37} \over 4}$<br><br>${{{a^2} + {b^2}} \over 8} + {{321} \over 4} - 81 = {{37} \over 4}$<br><br>${{{a^2} + {b^2}} \over 8} = 81 + {{37} \over 4} - {{321} \over 4}$<br><br>${{{a^2} + {b^2}} \over 8} = 81 - 71$<br><br>$\therefore$ a² + b² + 2ab = 144<br><br>80 + 2ab = 144<br><br>$\therefore$ 2ab = 64<br><br>$\therefore$ ${(a - b)^2} = {a^2} + {b^2} - 2ab = 80 - 64 = 16$